Question

In a triangle \(\mathrm{ABC}, \cos \alpha=\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}, \cos \beta=\frac{\mathrm{b}}{\mathrm{c}+\mathrm{a}}, \cos \gamma=\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}\) and \(\mathrm{k}=\tan ^{2} \frac{\alpha}{2} \cdot \tan ^{2} \frac{\beta}{2} \cdot \tan ^{2} \frac{\gamma}{2} .\) Then, (a) \(27 \mathrm{k}>1\) (b) \(27 \mathrm{k} \leq 1\) (c) \(9 \mathrm{k}=1\) (d) \(8 \mathrm{k}=1\)

Short answer

In this problem, we are given a triangle with cosine expressions of its angles and the given value k, which is the product of the squares of half-angle tangents. Using various trigonometric identities and theorems, we derive relationships between these expressions and compare them with the given options. We found that the option (b) \(27k \leq 1\) is true.

Step-by-step solution

Analyze the given expressions

We are given the cosine expressions for the angles of the triangle: $$\cos(\alpha) = \frac{a}{b+c}; \cos(\beta) = \frac{b}{c+a}; \cos(\gamma) = \frac{c}{a+b}$$ and the expression for k: $$k = \tan^2(\frac{\alpha}{2}) \cdot \tan^2(\frac{\beta}{2}) \cdot \tan^2(\frac{\gamma}{2})$$ The goal is to find the relationship between these expressions and compare them with the given options.

Apply the Sum of Angles in a triangle and Cosine Rule

We know that the sum of angles of any triangle is \(180^\circ\) or \(\pi\) radians. So, $$\alpha + \beta + \gamma = \pi$$ Applying the cosine rule to triangle ABC, we have: $$a^2 = b^2 + c^2 - 2bc \cos(\alpha)$$ $$b^2 = c^2 + a^2 - 2ca \cos(\beta)$$ $$c^2 = a^2 + b^2 - 2ab \cos(\gamma)$$

Substitute the given cosine expressions into Cosine Rule

We will substitute the cosine expressions given in the problem into the cosine rule: $$a^2 = b^2 + c^2 - 2bc \frac{a}{b+c}$$ $$b^2 = c^2 + a^2 - 2ca \frac{b}{c+a}$$ $$c^2 = a^2 + b^2 - 2ab \frac{c}{a+b}$$

Manipulate and simplify the equations from previous step

Now, let's manipulate and simplify these equations. Equation 1: $$a^2 = b^2 + c^2 - \frac{2abc}{b+c}$$ $$a^2 (b+c) = b^2(b+c) + c^2(b+c) - 2abc$$ $$ab^2 + ac^2 = ab^2 + bc^2 + cb^2 + c^2b - 2abc$$ $$ab^2 + ac^2 = ab^2 + bc^2 + bc^2 + ac^2 - 2abc$$ $$2abc = ab^2 + 2bc^2 + ac^2$$ (*) Similarily, we obtain from Equation 2 and Equation 3: $$2bca = b^2c + 2ca^2 + b^2a$$ (*) $$2cab = bc^2 + 2ab^2 + ca^2$$ (*)

Add the obtained equations from the previous step

Now, let's add the three obtained equations (*) together: $$2abc + 2bca + 2cab = ab^2 + 2bc^2 + ac^2 + b^2c + 2ca^2 + b^2a + bc^2 + 2ab^2 + ca^2$$ For the right-hand side of the equation, note that each of the following terms appears twice: \(ab^2, 2bc^2, ac^2\). Thus, we can simplify as follows: $$2abc + 2bca + 2cab = 2(ab^2 + bc^2 + ca^2)$$

Divide both sides of the equation and use substitution

Now, divide both sides of the equation by \(2abc\): $$abc + bca + cab = ab^2 + bc^2 + ca^2$$ $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{b^2 + bc + c^2}{abc}$$ Now, substitute the given cosine expressions back in: $$\frac{a+b+c}{c} = \frac{b^2 + bc + c^2}{abc}$$

Use double angle tangent formula

Recall that we are given: $$k = \tan^2(\frac{\alpha}{2}) \cdot \tan^2(\frac{\beta}{2}) \cdot \tan^2(\frac{\gamma}{2})$$ Using double angle tangent formula, we find: $$\tan^2(\frac{\alpha}{2}) = \frac{1 - \cos(\alpha)}{1 + \cos(\alpha)}$$ So, we have: $$k = \frac{1 - \cos(\alpha)}{1 + \cos(\alpha)} \cdot \frac{1 - \cos(\beta)}{1 + \cos(\beta)} \cdot \frac{1 - \cos(\gamma)}{1 + \cos(\gamma)}$$ Now substitute the given cosine expressions into this equation.

Solve for k and compare with the given options

After substituting the given cosine expressions and simplifying, we have: $$k = \frac{(b+c-a)(c+a-b)(a+b-c)}{(b+c+a)^3}$$ Therefore, option (b) \(27k \leq 1\) is true. So the final answer is: (b) \(27k \leq 1\).

Key concepts

Cosine Rule

When dealing with triangles, the Cosine Rule is a powerful tool used to find the lengths of sides or angles. The rule is particularly helpful for triangles that do not have a right angle. The cosine rule relates the lengths of the sides of a triangle to the cosine of one of its angles.

For triangle \( ABC \), where \( a, b, \) and \( c \) are the sides opposite angles \( \alpha, \beta, \) and \( \gamma \) respectively, the Cosine Rule is given by:

• \( a^2 = b^2 + c^2 - 2bc \cdot \cos(\alpha) \)
• \( b^2 = a^2 + c^2 - 2ac \cdot \cos(\beta) \)
• \( c^2 = a^2 + b^2 - 2ab \cdot \cos(\gamma) \)

These formulas allow us to calculate an unknown side when we know two sides and the included angle. They are also useful to determine an angle when all three sides are known.

In the given exercise, we used cosine relationships to understand how the sides and internal angles of the triangle interact. We substituted expressions simplified from given cosine values into the cosine rule equations to derive relationships useful for solving the problem.

Triangle Angle Properties

Triangles have distinctive properties, particularly concerning their internal angles. One crucial property is the sum of internal angles, which is always \( 180^\circ \) or \( \pi \) radians. This property holds true no matter the type of triangle—be it scalene, isosceles, or equilateral.

In our problem, this property is key because it allowed us to establish the relationship \( \alpha + \beta + \gamma = \pi \). Recognizing this helps in simplifying equations or checking the correctness of results when solving for unknown angles.

Understanding how to maneuver between different trigonometric identities with the context of this property helps in simplifying complex expressions in problems like those in trigonometry exercises.

Tangent Formula in Triangles

The tangent formula in triangles is another vital tool, often used in determining angles and side relationships. In trigonometry, the tangent of half an angle has a special form, given by:

• \( \tan\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1 - \cos(\alpha)}{1 + \cos(\alpha)}} \)

This formula is especially useful for deriving expressions involving the angles divided by two.

The exercise used this to form an expression for \( k \), where \( k = \tan^2\left(\frac{\alpha}{2}\right) \cdot \tan^2\left(\frac{\beta}{2}\right) \cdot \tan^2\left(\frac{\gamma}{2}\right) \). By substituting in the known \( \cos(\alpha), \cos(\beta), \) and \( \cos(\gamma) \), the problem cleverly reduced complex trigonometric relationships into a simpler form.

Mastering how to use and manipulate these tangent formulas is crucial for solving advanced trigonometry problems in triangles, providing a gateway to deeper understanding and application in different mathematical contexts.